H-Index

Given an array of citations (each citation is a non-negative integer) of a researcher, write a function to compute the researcher’s h-index.

According to the definition of h-index on Wikipedia: “A scientist has index h if h of his/her N papers have at least h citations each, and the other N − h papers have no more than h citations each.”

For example, given citations = [3, 0, 6, 1, 5], which means the researcher has 5 papers in total and each of them had received 3, 0, 6, 1, 5 citations respectively. Since the researcher has 3 papers with at least 3 citations each and the remaining two with no more than 3 citations each, his h-index is 3.

Note: If there are several possible values for h, the maximum one is taken as the h-index.

Hint:

An easy approach is to sort the array first.
What are the possible values of h-index?
A faster approach is to use extra space.

【题意】
给定研究人员的文章引用次数的数组（每一篇文章的引用次数都是非负整数），编写函数计算该研究人员的h指数。

根据维基百科对h指数的定义：“一名科学家的h指数是指其发表的N篇论文中，有h篇论文分别被引用了至少h次，其余N-h篇的引用次数均不超过h次”。

例如，给定引用次数数组 = [3, 0, 6, 1, 5]，这意味着研究人员总共有5篇论文，每篇分别获得了3, 0, 6, 1, 5次引用。由于研究人员有3篇论文分别至少获得了3次引用，且其余两篇的引用次数不超过3次，因而其h指数是3。

注意：如果存在多个可能的h值，取最大值作为h指数。
【算法】
从小到大排序后，指针指向末位，维护一个count变量，如果数组中的值大于等于count，则count++。指针向前继续移动，因为数组中数越来越小，所以大于count就能保证后边也大于count。

【code】

class Solution {public:    int hIndex(vector<int>& citations) {        sort(citations.begin(),citations.end());        int count = 0;        for (int i = citations.size()-1; i >= 0; i--)        {            if(citations[i] > count+1)                count++;            else                break;        }        return count;    }};